TSTP Solution File: SYN944+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN944+1 : TPTP v8.1.2. Released v3.1.0.
% Transfm : none
% Format : tptp:raw
% Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% Computer : n029.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Fri Sep 1 03:36:36 EDT 2023
% Result : Theorem 0.14s 0.37s
% Output : Proof 0.14s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SYN944+1 : TPTP v8.1.2. Released v3.1.0.
% 0.00/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34 % Computer : n029.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Sat Aug 26 22:05:55 EDT 2023
% 0.14/0.34 % CPUTime :
% 0.14/0.37 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.14/0.37
% 0.14/0.37 % SZS status Theorem
% 0.14/0.37
% 0.14/0.38 % SZS output start Proof
% 0.14/0.38 Take the following subset of the input axioms:
% 0.14/0.38 fof(prove_this, conjecture, ![A, B, C]: ((s(A) & (s(B) & (r(B, C) & (![X]: (s(X) => p(X)) & ![Y, X2]: (r(X2, Y) => q(X2, Y)))))) => ?[X2, Y2]: (p(X2) & q(X2, Y2)))).
% 0.14/0.38
% 0.14/0.38 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.14/0.38 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.14/0.38 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.14/0.38 fresh(y, y, x1...xn) = u
% 0.14/0.38 C => fresh(s, t, x1...xn) = v
% 0.14/0.38 where fresh is a fresh function symbol and x1..xn are the free
% 0.14/0.38 variables of u and v.
% 0.14/0.38 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.14/0.38 input problem has no model of domain size 1).
% 0.14/0.38
% 0.14/0.38 The encoding turns the above axioms into the following unit equations and goals:
% 0.14/0.38
% 0.14/0.38 Axiom 1 (prove_this_1): s(b) = true2.
% 0.14/0.38 Axiom 2 (prove_this_2): r(b, c) = true2.
% 0.14/0.38 Axiom 3 (prove_this_3): fresh(X, X, Y) = true2.
% 0.14/0.38 Axiom 4 (prove_this_3): fresh(s(X), true2, X) = p(X).
% 0.14/0.38 Axiom 5 (prove_this_4): fresh2(X, X, Y, Z) = true2.
% 0.14/0.38 Axiom 6 (prove_this_4): fresh2(r(X, Y), true2, X, Y) = q(X, Y).
% 0.14/0.38
% 0.14/0.38 Goal 1 (prove_this_5): tuple(p(X), q(X, Y)) = tuple(true2, true2).
% 0.14/0.38 The goal is true when:
% 0.14/0.38 X = b
% 0.14/0.38 Y = c
% 0.14/0.38
% 0.14/0.38 Proof:
% 0.14/0.38 tuple(p(b), q(b, c))
% 0.14/0.38 = { by axiom 6 (prove_this_4) R->L }
% 0.14/0.38 tuple(p(b), fresh2(r(b, c), true2, b, c))
% 0.14/0.38 = { by axiom 2 (prove_this_2) }
% 0.14/0.38 tuple(p(b), fresh2(true2, true2, b, c))
% 0.14/0.38 = { by axiom 5 (prove_this_4) }
% 0.14/0.38 tuple(p(b), true2)
% 0.14/0.38 = { by axiom 4 (prove_this_3) R->L }
% 0.14/0.38 tuple(fresh(s(b), true2, b), true2)
% 0.14/0.38 = { by axiom 1 (prove_this_1) }
% 0.14/0.38 tuple(fresh(true2, true2, b), true2)
% 0.14/0.38 = { by axiom 3 (prove_this_3) }
% 0.14/0.38 tuple(true2, true2)
% 0.14/0.38 % SZS output end Proof
% 0.14/0.38
% 0.14/0.38 RESULT: Theorem (the conjecture is true).
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